| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\-2\\3\end{pmatrix} - \begin{pmatrix}-2\\-3\\-5\end{pmatrix} = \begin{pmatrix}4\\1\\8\end{pmatrix}\) | B1 | Finds direction of one line to another |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&3&5\\2&-3&1\end{vmatrix} = \begin{pmatrix}18\\14\\6\end{pmatrix} \sim \begin{pmatrix}9\\7\\3\end{pmatrix}\) | M1 A1 | Find common perpendicular |
| \(\frac{1}{\sqrt{139}}\begin{vmatrix}\begin{pmatrix}4\\1\\8\end{pmatrix}\cdot\begin{pmatrix}9\\7\\3\end{pmatrix}\end{vmatrix} = \frac{67}{\sqrt{139}}\ (=5.68)\) | M1 A1 | Uses formula for shortest distance |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&0&1\\-4&3&5\end{vmatrix} = \begin{pmatrix}3\\9\\-3\end{pmatrix} \sim \begin{pmatrix}1\\3\\-1\end{pmatrix}\) | M1 A1 | Finds vector perpendicular to the plane |
| \(1(-1)+3(-3)-1(-4) = -6 \Rightarrow x+3y-z=-6\) | M1 A1 | Uses point in the plane |
| Total: 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\-2\\3\end{pmatrix} - \begin{pmatrix}-2\\-3\\-5\end{pmatrix} = \begin{pmatrix}4\\1\\8\end{pmatrix}$ | B1 | Finds direction of one line to another |
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&3&5\\2&-3&1\end{vmatrix} = \begin{pmatrix}18\\14\\6\end{pmatrix} \sim \begin{pmatrix}9\\7\\3\end{pmatrix}$ | M1 A1 | Find common perpendicular |
| $\frac{1}{\sqrt{139}}\begin{vmatrix}\begin{pmatrix}4\\1\\8\end{pmatrix}\cdot\begin{pmatrix}9\\7\\3\end{pmatrix}\end{vmatrix} = \frac{67}{\sqrt{139}}\ (=5.68)$ | M1 A1 | Uses formula for shortest distance |
| **Total: 5** | | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&0&1\\-4&3&5\end{vmatrix} = \begin{pmatrix}3\\9\\-3\end{pmatrix} \sim \begin{pmatrix}1\\3\\-1\end{pmatrix}$ | M1 A1 | Finds vector perpendicular to the plane |
| $1(-1)+3(-3)-1(-4) = -6 \Rightarrow x+3y-z=-6$ | M1 A1 | Uses point in the plane |
| **Total: 4** | | |
---4 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations
$$\mathbf { r } = - 2 \mathbf { i } - 3 \mathbf { j } - 5 \mathbf { k } + \lambda ( - 4 \mathbf { i } + 3 \mathbf { j } + 5 \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 2 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } + \mu ( 2 \mathbf { i } - 3 \mathbf { j } + \mathbf { k } )$$
respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.\\
The plane $\Pi$ contains $l _ { 1 }$ and the point with position vector $- \mathbf { i } - 3 \mathbf { j } - 4 \mathbf { k }$.
\item Find an equation of $\Pi$, giving your answer in the form $a x + b y + c z = d$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q4 [9]}}